Your strategy is effectively a binary search. Assuming a perfect distribution (it is always possible to find a letter that is in exactly half of the remaining words), you should be able to guess (on average) up to twice as many times as you are allowed wrong guesses before losing the game. Of course, the last guess still has a 50% chance of being wrong so I believe your strategy is guaranteed to work when the total number of initial candidates is 2^^{(2G - 1)} or less, where G represents the number of allowed wrong guesses. For instance, when G = 7 then you are guaranteed (100%) to win when the initial number of candidates is 8192 or less and 50% when it is up to 16,384. For eclectic, there are 9638 initial words in my word list so only a 50% chance of success.

My strategy is optimized not to guess wrong. After only 5 guesses (2 right and 3 wrong) it had narrowed the search space down to exactly 1 word. This is because I prune by position not just by presence of letter so even successful guesses on a popular letter can still effectively decrease the search space. Your approach would be improved with this strategy as well. I don't think the opportunities stop there.

In a private /msg with Lawliet the idea of finding the solution with the least number of guesses was proposed. I indicated that my strategy would change - but to what? A binary search is already optimal. There needs to be some balance then between improving your odds from 50/50 of guessing wrong while still effectively reducing your search space each time. This way you can survive long enough to win but not guess ad nauseum.

I am kicking around some ideas where you still look for a very popular letter (say in 70% of the remaining words) but by position would still split that 70% in half if you guessed right. The result then would be that you would guess wrong 30% of the time and remove 70% of your search space or guess right 70% of the time but reduce your search space by 65%. Does this sound viable to you?